linux - PHP 的 SELECT 语句问题

Question

+--------------+------+
| IP           | Say  |
+--------------+------+
| 192.168.1.1  |    1 |
+--------------+------+    

$con = mysqli_connect("$host", "$user", "$pass", "$db_name") or die("cannot connect");
$Q1 = "SELECT Say From spam_engel WHERE IP = '192.168.1.1'";
$ol = mysqli_query($con, $q1);
echo gettype($ol);

打印的东西是“NULL”。但它应该打印“int”......可能是什么问题？

Answer 1

PHP variables区分大小写:

Variables in PHP are represented by a dollar sign followed by the name of the variable. The variable name is case-sensitive.

所以在你的代码中:

$Q1 = "SELECT Say From spam_engel WHERE IP = '192.168.1.1'";
$ol = mysqli_query($con, $q1);

在这里你使用 $Q1 然后传递给你的 mysql_query() 小写变量 $q1;

要解决此问题，请统一变量名的大小写，如下所示:

$q1 = "SELECT Say From spam_engel WHERE IP = '192.168.1.1'";
$ol = mysqli_query($con, $q1);